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The differential gear of a car axle allows the wheel on the left side of a car to rotate at a different angular speed than the wheel on the right side. A car is driving at a constant speed around a circular track on level ground, completing each lap in 19.5 s. The distance between the tires on the left and right sides of the car is 1.60 m, and the radius of each wheel is 0.350 m. What is the difference between the angular speeds of the wheels on the left and right sides of the car?

Short Answer

Expert verified
The difference in angular speeds is approximately 0.73 rad/s.

Step by step solution

01

Understand the problem

We need to find the difference in angular speeds between the left and right wheels of a car as it travels around a circular track, given the time per lap, and the distance and radius measurements of the wheels.
02

Determine the circumference of the circular track

Identify that the car completes one full circle in 19.5 seconds. Since we don't have the track's radius, we'll work in terms of the circumference: let it be denoted as \( C \).
03

Establish wheel path circumferences

The wheels travel in separate paths; the left wheel travels around the inner circle and the right around the outer circle. These circles differ by the distance between the tires, 1.60 m.
04

Express inner and outer circumferences

Let the inner circumference, the path for the left wheel, be \( C_{inner} = C - 1.6 \cdot \pi \) and the outer circumference for the right wheel be \( C_{outer} = C + 1.6 \cdot \pi \) using \( \, \pi \approx 3.14159 \).
05

Compute angular speeds

The linear speed equals the circumference traveled divided by time, hence:- \( v_{inner} = \frac{C_{inner}}{19.5} \)- \( v_{outer} = \frac{C_{outer}}{19.5} \)Since \( \omega = \frac{v}{r} \), where \( r \) is the wheel radius \( 0.350 \, \text{m} \), calculate:- \( \omega_{left} = \frac{v_{inner}}{0.350} \)- \( \omega_{right} = \frac{v_{outer}}{0.350} \)
06

Calculate the difference in angular speeds

The difference in angular speeds is \( \Delta \omega = \omega_{right} - \omega_{left} \). Substitute the expressions from Step 5 and calculate:\[ \Delta \omega = \frac{C_{outer} - C_{inner}}{19.5 \times 0.350} = \frac{1.6 \cdot \pi}{19.5 \times 0.350} \]
07

Perform the calculation

Plug in the numbers to find:\[ \Delta \omega = \frac{1.6 \cdot 3.14159}{19.5 \times 0.350} \approx 0.73 \, \text{rad/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Gear
The differential gear is a critical part of a car's drivetrain, allowing for different rotational speeds between the left and right wheels. This mechanism is vital for maintaining traction and stability when a car navigates curves. Every time you drive through a curve, the inner wheel must cover a shorter distance than the outer wheel. If both wheels rotated at the same speed, the wheels would skid, causing wear on tires and potentially losing control.
The differential gear compensates for this by splitting the torque so that each wheel rotates at an appropriate speed depending on the curve. By allowing for different angular speeds, the differential gear ensures smooth turns and enhances the driving experience. Understanding its function helps in grasping how vehicles efficiently manage circular track motion.
Circular Track Motion
When a car moves along a circular track, it engages in circular motion. This means every part of the car is continuously changing direction, requiring careful coordination of speed and steering. The car's inner side takes a path smaller in radius compared to the outer side, affecting how fast each side has to move.
In our exercise, knowing the car completes the track in a specific time helps us compute how much distance each wheel covers. The time duration ensures that we account for the circular nature, allowing for precise calculations of paths and speeds. Comprehending circular track motion lays the groundwork for understanding how differential gears and angular velocities play roles in vehicle dynamics.
Linear Speed Calculation
Linear speed is crucial in determining how fast a wheel is moving along its path. For a body in motion, linear speed is the distance traveled divided by the time taken. In circular motion, linear speed converts into tangential speed, helping to calculate the angular velocity of a rotating wheel.
For our scenario, we translate the path circumference into linear speed by dividing by the time per lap. For instance, if one wheel covers the inner path, its speed calculation involves the smaller circumference, whereas the outer wheel uses a larger one. By determining each wheel's linear speed, we bridge to understanding their differing angular velocities.
Angular Velocity
Angular velocity represents the speed of rotation and is a fundamental concept in understanding rotational dynamics. It describes how fast an object rotates or revolves relative to another point. In this exercise, it's crucial for calculating the difference in speed between the two car wheels.
We derive angular velocity by using the formula \( \omega = \frac{v}{r} \), where \( v \) is linear speed and \( r \) is the radius of the wheel. The differential in these velocities stems from differing path circumferences, dictated by the vehicle's differential gear. Once we compute each wheel's angular velocity, we can deduce their difference, illustrating why angular velocity plays a pivotal role in mechanics.
Wheel Path Circumference
Wheel path circumference is a measure of distance each wheel travels during a lap around the track. Understanding this concept helps differentiate the paths taken by the left and right wheels because of their placement relative to the track center.
In our exercise, the path for the left wheel, being the inner path, is smaller than that of the right wheel. We express these paths in terms of circumference as: \( C_{inner} = C - 1.6 \cdot \pi \) and \( C_{outer} = C + 1.6 \cdot \pi \). These equations account for the width of the car (1.60 m) and illustrate how physical dimensions affect the wheel speeds. Grasping how path circumferences develop aids in solving rotational dynamics problems.
Physics Problem Solving
Solving physics problems often involves breaking down a scenario into understandable parts. This structured approach helps tackle complex situations systematically. Here's how we applied it to our car and track problem:
  • Problem Understanding: Recognize the need to determine angular speed differences.
  • Identify Given Data: Confirm what measurements and time values are provided.
  • Establish Relationships: Define how wheel paths' circumferences change with the car's dimensions.
  • Use Formulas: Relate linear speed to angular velocity for each wheel.
  • Calculate: Execute numerical calculations precisely for the desired result using derived formulas.
Mastering these steps is crucial for effective physics problem solving, revealing the real-world applications of theoretical concepts.

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Most popular questions from this chapter

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